# two positive point charges of 12uc and 8uc are 10cm apart. the work done in bringing them 4cm closer is?

Sourabh Singh IIT Patna
8 years ago
Hii

Find out the change in the potential energy of the system and take the difference. You will get the answer.

Best of Luck
11 Points
6 years ago
q1=12×10^-6,q2=8×10^-6.
Wkt,potential when the are at R1(10 cm)
V1=kq1q2/R1=86.4×10^-1~=86×10^-1.
V2=kq1q2/R2=216×10^-1.
W.D=V2-V1=(216×10^-1)-(86×10^-1)=130×10^-1
=13J
15 Points
6 years ago
So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta
4 years ago

So the primary distance r1=10cm=0.1m
Final distance= 10 - 4=6cm=0.06m
Now the workdone=9×10^9 × 8×10^-6 ×12×10^-6(1/0.06-1/0.1)=5.76J
Krish Gupta
4 years ago
HiiFind out the change in the potential energy of the system and take the difference. You will get the answer.Best of Luck

4 years ago
Dear student,

Initial distance between the charges, d1 = 10 cm = 0.1 m
Final distance between the charges, d2 = 10 cm – 4 cm = 6 cm = 0.06 m
Hence, W = 9 x 109 x 12 x 10-6 x 8 x 10-6 (1/0.06 – 1/0.1)
= 5.76 J

Hope it helps.
Thanks and regards,
Kushagra